Question

If triangle ABC is an isosceles triangle in which AB = AC = 13 cm then find the value of $\frac{\text{area of}\Delta ADC}{\text{area of}\Delta EFB}.$

• A. $\frac{165}{320}$
• B. $\frac{13}{18}$
• C. $\frac{150}{319}$
• D. $\frac{169}{324}$

Answer 1

The correct option is D

$\frac{169}{324}$

In $\Delta ADC$ and $\Delta FEB,$

$\angle C=\angle B$

(since base angles of isosceles triangle are equal)

$\angle ADC=\angle EFG={90}^{\circ }$

$\therefore \Delta ADC\sim \Delta EFB$ (by AA similarity criterion)

$\Rightarrow \frac{\text{area of}\Delta ADC}{\text{area of}\Delta EFB}=\frac{A{C}^2}{B{E}^2}=\frac{A{C}^2}{(BC+CE{)}^2}$

(The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides)

$\Rightarrow \frac{\text{area of}\Delta ADC}{\text{area of}\Delta EFB}=\frac{(13{)}^2}{(10+8{)}^2}$

$=\frac{(13{)}^2}{(18{)}^2}=\frac{169}{324}$